Multiply the following complex numbers: $({3-3i}) \cdot ({4-2i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3-3i}) \cdot ({4-2i}) = $ $ ({3} \cdot {4}) + ({3} \cdot {-2}i) + ({-3}i \cdot {4}) + ({-3}i \cdot {-2}i) $ Then simplify the terms: $ (12) + (-6i) + (-12i) + (6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 12 + (-6 - 12)i + 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 12 + (-6 - 12)i - 6 $ The result is simplified: $ (12 - 6) + (-18i) = 6-18i $